I love floor function

Algebra Level 5

Let $S$ denote the number of solutions that satisfy the equation $\left \lfloor \frac x{1001} \right \rfloor = \frac x{1003}$ . Find $\left \lfloor \frac S {67} \right \rfloor$ .

The answer is 7.

2 solutions

Rishi Sharma
Apr 11, 2015

$Since\quad L.H.S\quad have\quad a\quad floor\quad function\quad \\ So\quad R.H.S\quad must\quad also\quad be\quad an\quad integer.\\ So\quad x\quad must\quad be\quad a\quad multiple\quad of\quad 1003.\quad \\ Replace\quad x\quad by\quad 1003n\quad where\quad n\quad is\quad any\quad integer.\\ So\quad our\quad equation\quad becomes\quad \left\lfloor \frac { 1003n }{ 1001 } \right\rfloor =n\\ Now,\quad using\quad \left\lfloor x \right\rfloor =x-\left\{ x \right\} \quad \\ (here\quad \left\{ x \right\} \quad represents\quad fractional\quad part\quad of\quad x)\\ We\quad have,\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \frac { 1003n }{ 1001 } -\left\{ \frac { 1003n }{ 1001 } \right\} =n\\ On\quad simplification\quad we\quad get,\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \frac { 2n }{ 1001 } =\left\{ \frac { 1003n }{ 1001 } \right\} \\ Now\quad using\quad 0\le \left\{ x \right\} <1,\quad We\quad get\quad 0\le n<500.5\quad \\ \therefore \quad n\quad can\quad take\quad 501\quad values.\quad Hence\quad S=501\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \left\lfloor \frac { 501 }{ 67 } \right\rfloor =7$

Very impressive. But I wonder, why do you exclude negative values of X? Aren't there also 500 negative values of X for which this equality holds true? (-1003, -2006, -3009, etc.?)

Paul Josephson - 6 years, 1 month ago

In my solution when I assumed n as a variable I have specified that n is any integer so when I finally found the range for n it naturally came out n>0. If you are not satisfied you can check by using a calculator like for -1003 the L.H.S will be -2 where else R.H.S will be -1. Same will be true for other values also. If you are still not sure use desmos (graphing calculator) put the equation in the form floor(x/a)=x/b (b>a)where a and b can change values. You will notice that there are no real solutions possible for x<0. If you still have any other doubt please tell me.

Rishi Sharma - 6 years, 1 month ago

the floor is defined as follows for any $x \in \mathbb{R}$ : it is the unique integer $k$ such that $k \leqslant x < k+1$ . From there you can see that the negative values should not be added (because for $k>0$ , $-k\frac{1003}{1001} < -k$ , so we never have this equality !)

Charles Dutertre - 6 years, 1 month ago

In my solution to this equation(I have attached a photo in my solution) , I deal with 2 inequalities at the same time and the solution to this is the intersection of solution set of both the inequalities.From first inequality,(refer inequality "a") we get to the conclusion that any integer to be the solution of the equation has to be non-negative (so 0 is also a solution but no negative numbers)

Dheeman Kuaner - 6 years ago

Exactly the same

Aakash Khandelwal - 5 years, 1 month ago

Dheeman Kuaner
May 21, 2015

I have attached my solution in the form of image file

I love floor function

The answer is 7.

2 solutions

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